College Algebra, Chapter 8, 8.4, Section 8.4, Problem 30

Complete the square to determine whether the equation represents an ellipse, a parabola, a hyperbola, or a degenerate conic. If the graph is an ellipse, find the center, foci, vertices and lengths of the major and minor axes. If it is a parabola, find the vertex, focus and directrix. If it is a hyperbola, find the center, foci, vertices and asymptotes. Sketch the graph of the equation. If the equation has no graph, explain why.


$\begin{equation} \begin{aligned} 4x^2 - 4x - 8y + 9 =& 0 && \text{Subtract } 9 \\ \\ 4x^2 - 4x - 8y =& -9 && \text{Factor and Group terms} \\ \\ 4(x^2 - x) - 8y =& -9 && \text{Complete the square: add } \left( \frac{-1}{2} \right)^2 = \frac{1}{4} \text{ on the left side and $1$ on the right side} \\ \\ 4 \left(x^2 - x + \frac{1}{4} \right) - 8y =& -9 + 1 && \text{Perfect square} \\ \\ 4 \left(x - \frac{1}{2} \right)^2 - 8y =& -8 && \text{Add } 8y \\ \\ 4 \left( x - \frac{1}{2} \right)^2 =& 8y - 8 && \text{Divide by } 4 \\ \\ \left(x - \frac{1}{2} \right)^2 =& 2y - 2 && \text{Factor } 2 \\ \\ \left(x - \frac{1}{2} \right)^2 =& 2 (y - 1) && \end{aligned} \end{equation}$


The equation is a parabola that opens upward with vertex at $\displaystyle \left( \frac{1}{2}, 1 \right)$. It is obtained from the parabola $x^2 = 2y$ by shifting it $\displaystyle \frac{1}{2}$ units to the right and $1$ unit upward. Since $4p = 2$, we have $\displaystyle p = \frac{1}{2}$. So the focus is $\displaystyle \frac{1}{2}$ units above the vertex and the directrix is $\displaystyle \frac{1}{2}$ units below the vertex.

Therefore, the focus is at

$\displaystyle \left( \frac{1}{2}, 1 \right) \to \left( \frac{1}{2}, 1 + \frac{1}{2} \right) = \left( \frac{1}{2}, \frac{3}{2} \right)$

and the directrix is the line

$\displaystyle y = 1 - \frac{1}{2} = \frac{1}{2}$